Data Sufficiency

I need help!

If x and y are positive, is 4x>3y?

1. x>y-x

2. x/y<1

I get how the two statements alone are not sufficient but how about together??

this is what i did so far:

1. x>y-x becomes 2x>y some values that satisfy this are x=1, y=1....x=2, y=3.... x=3, y =5

putting these values in 4x>3y gives true/false statements....2>1 yes......8>9...no..12>15..no

so insufficient

2. since x, >0 x/y <1 becomes x<y

values of x,y that satisfy this are...4,2...6,1...3,4

putting these in 4x>3y

we get Y/N statements agn.....16>6 yes.....24>3 yes.....12>12...no

so insufficient

BUT together...i'm not sure how to combine the two statements:

2x>y & x<y to see if 4x> 3y

What do i do??? It would take me forever to find values to satisfy both stat 1 & 2 and then to check if any of the values give a consistent answer for 4x>3y. Is there a shortcut that i'm missing?????

Also another issue related to this i have is that i never know which "trial" values to pick. I tend to start with small positive no.s then i try higher positive no.s then if the question allows i'll plug in negative values or fractions.


my question is how do i speed up this process? any tips on plug-in values would be greatly appreciated! thanks

>>>

raz1024 wrote:Hey Mitz (?),

Nice job in getting everything down to 2x>y & x<y.

Combining those two statements, you can form a single inequality:

x<y<2x

Now, you're interested in 3y, according to the prompt, so multiply everything by 3:

3x<3y<4x. ! shouldnt this be 6x ?

You know from this inequality that 3y is less than 4x, and thus 4x>3y. SUFFICIENT.

As for plugging in values, it looks like you did just fine. Your strategy of starting with smaller numbers is exactly what should be done, because it keeps everything simple.

You could also think about it strictly in terms of the algebra:

(1) tells you that 2x>y, which means 4x>2y. Does that mean 4x>3y? Not necessarily.

(2) tells you x<y, which means 4x<4y. Does that mean 4x>3y? Not necessarily.

Thanks Raz1024!

Ok, so you're saying...

given 2x>y & x<y

we get x<y<2x

since we want to know the answer in relation 3y, multiply all by 3 so we get

3x<3y<6x

so since 6x>3y ....4x must be greater than 3y... correct?

BUT ! there's an issue. the answer provided is E not C. So maybe there's a flaw in there somewhere...?!

The answer should be E

-1 ) 2x > y --4x >2y
-2) x<y-->4x<4y

Combining both
2y<4x<4y

Even both statements do not tell us that 4x >3y. Hope that clarifies. Hence E.

The answer is indeed E...why? Because sometimes even instructors make silly mistakes

Nice job catching the error from yours truly!

So, the lesson here? Be careful with your calculations. Although at this point, my saying that is like the pot and the kettle. Do as I say, not as I do

raz1024 wrote:The answer is indeed E...why? Because sometimes even instructors make silly mistakes

Nice job catching the error from yours truly!

So, the lesson here? Be careful with your calculations. Although at this point, my saying that is like the pot and the kettle. Do as I say, not as I do

This famous saying must have been invented by a GMAT instructor only, very entertaining.

mitzwillrockgmat wrote:I need help!


2x>y & x<y to see if 4x> 3y

What do i do??? It would take me forever to find values to satisfy both stat 1 & 2 and then to check if any of the values give a consistent answer for 4x>3y. Is there a shortcut that i'm missing?????

Also another issue related to this i have is that i never know which "trial" values to pick. I tend to start with small positive no.s then i try higher positive no.s then if the question allows i'll plug in negative values or fractions.


my question is how do i speed up this process? any tips on plug-in values would be greatly appreciated! thanks

It won't take forever if you think about the concepts behind the plug ins. At this point, you are trying to show that 4x can be both greater and smaller than 3y, while satisfying both statements. You know that 2x is greater than y, but y is greater than x. So let's first try with an x and y that are very close to each other: x=1 and y=1.1, so that 2x=2 is greater than y=1.1.
in this case, 4x=4 is greater than 3y=3.3, so we have a yes.
Now think: Can we have a "no", preferably without reinventing the wheel altogether?
Keep x as 1, so that 4x is still 4. All we need is a y that is great enough to pass 3y>4: y=1.5 would work, and still satisfy the requirement of 2x>y.

Take home messages:
1) plug in with direction, not blindly: plug in one set, see what answer you get, then think about which plug in will get you the opposite answer.
2) When plugging in a second time for more than one value, keep one and play with the other.

You can also solve this question graphically as I have discussed on the Veritas blog here:
https://www.veritasprep.com/blog/2011/0 ... -part-iii/

mitzwillrockgmat wrote:I need help!

If x and y are positive, is 4x>3y?

1. x>y-x

2. x/y<1

I get how the two statements alone are not sufficient but how about together??

this is what i did so far:

1. x>y-x becomes 2x>y some values that satisfy this are x=1, y=1....x=2, y=3.... x=3, y =5

putting these values in 4x>3y gives true/false statements....2>1 yes......8>9...no..12>15..no

so insufficient

2. since x, >0 x/y <1 becomes x<y

values of x,y that satisfy this are...4,2...6,1...3,4

putting these in 4x>3y

we get Y/N statements agn.....16>6 yes.....24>3 yes.....12>12...no

so insufficient

BUT together...i'm not sure how to combine the two statements:

2x>y & x<y to see if 4x> 3y

What do i do??? It would take me forever to find values to satisfy both stat 1 & 2 and then to check if any of the values give a consistent answer for 4x>3y. Is there a shortcut that i'm missing?????

Also another issue related to this i have is that i never know which "trial" values to pick. I tend to start with small positive no.s then i try higher positive no.s then if the question allows i'll plug in negative values or fractions.


my question is how do i speed up this process? any tips on plug-in values would be greatly appreciated! thanks

mitzwillrockgmat wrote:I need help!

If x and y are positive, is 4x>3y?

1. x>y-x

2. x/y<1

Since y is POSITIVE, the inequalities can safely be divided by y.

4x > 3y
x > (3/4)y
x/y > 3/4
Question stem, rephrased:
Is x/y > 3/4?

Statement 1:
x > y-x
2x > y
2(x/y) > 1
x/y > 1/2

Case 1: x/y = 2/3
In this case, x/y < 3/4, so the answer to the rephrased question stem is NO.
Case 2: x/y = 9/10
In this case, x/y > 3/4, so the answer to the rephrased question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Cases 1 and 2 satisfy BOTH statements.
Since the answer is NO in Case 1 but YES in Case 2, the two statements combined are INSUFFICIENT.

The correct answer is E.